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The thermochemical equation for the combustion of butane is: $$\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(l)} , \quad\Delta H = -5748 \ \mathrm{kJ\cdot mol^{−1}}$$

So far, I've calculated the heat released by water as follows:

$$\begin{align}Q&= m \times c \times \Delta\vartheta\\ &= \rm 0.724\ g \times 4.18 \times 43.2 \\ &= \rm 130.7\ \frac{kJ}{mol}\end{align}$$

I don't know how to carry on from here.


marked as duplicate by Loong Jun 17 '18 at 17:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Firstly, you need to understand what is happening. Butane gas is burnt which releases heat energy. This heat energy is absorbed by water. As the water absorbs this heat energy, it increases in temperature from $\mathrm{7.44}$ to $\mathrm{50.7}$ degrees Celsius. So in your in working out, the formula you have used is correct but you have written the 'heat released by water' which is incorrect. The water absorbs heat energy which is why it increases in temperature.

Secondly, always remember to write down the units when you are using equations. That way you won't make silly mistakes and always end up with the correct units. For example, in your working out you got that the heat energy absorbed was $\mathrm{130.7\ kJ\ mol^{-1}}$. It should be clear that this the wrong unit as energy is measured in joules, not joules per a mole.

Now, the first step in doing this question is to calculate the energy that is required to heat the water up to the desired temperature: $$q = mc\Delta T$$

$$= \mathrm{724\ g\times4.18\ J\ g^{-1}\ K^{-1}\times43.26\ K = 130918.6\ J = 130.9\ kJ}$$

Note how the final answer in joules which is the correct unit (also my answer differs slightly to yours because for $\Delta T$ I used $\mathrm{43.26}$ rather than $\mathrm{43.2}$)

So now we know that we need $\mathrm{130.9\ kJ}$ of energy to heat the water up. This means that the burning of butane must release $\mathrm{130.9\ kJ}$ of energy as $\Delta H = q$.

By looking at the equation that is given, it can be seen that burning 1 mole of butane releases $\mathrm{5748\ kJ}$ of energy (you can tell it releases energy as $\Delta H < 0$ ). So to calculate the number of moles of butane required: $$n_\text{butane} = \mathrm{\frac{130.9\ kJ}{5748\ kJ\ mol^{-1}} = 0.023\ mol}$$

Note you know that you have used the right equation as the units that we got for the final answer is in moles which is correct

Now that we know the number of moles of butane that is required, all we have to do is convert it into grams: $$m_\text{butane} = \mathrm{0.023\ mol\times58.12\ g\ mol^{-1} = 1.32\ g}$$

Therefore your final answer is: 1.32 g

Note your final answer should be to 3 significant figures as the data in the question is to 3 significant figures


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